1089 Trick
Classic Puzzle
Take a three-digit number (one where the first and last digits are different by at least $2$).
Reverse it, and get the difference between that and the first number.
Reverse that, and add that to the difference you calculated just now.
You get $1089$.
Proof
Let the number you started with be expressed in decimal notation as $[abc]_{10}$.
Then it is $10^2 a + 10 b + c$.
Its reversal is:
- $10^2 c + 10 b + a$
The difference between the two is $99a - 99c$ (or $99c - 99a$, it matters not).
This is a multiple of $99$.
The three-digit multiples of $99$ are:
- $198$
- $297$
- $396$
- $495$
- $594$
- $693$
- $792$
- $891$
By adding any one of these to its reversal, you get:
- $9 \times 100 + 2 \times 9 \times 10 + 9 = 1089$
$\blacksquare$
Note: You need to make sure the difference between the first and last digits of the number you started with is at least $2$ so as to make sure that the first difference you calculate is definitely a $3$-digit number.
Sources
- 1986: David Wells: Curious and Interesting Numbers ... (previous) ... (next): $1089$
- 1995: Terry Wiley: Surreal School Stories: No. $2$ Vol. $1$: The Late Seventies Masonettes: The Second Chapter: 'My Brain Hurts'
- Jocasta Dribble has been passed the list of number codes for the secret society she has just joined. The code for "weird" is $1089$: "a funny number". Examples of its use ($234$, $742$ and $876$) are illustrated on the piece of paper.
- 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $1089$
- 2002: David Acheson: 1089 and all that